⇒Trigonometric Functions

Correct answer option is (D)

The given equation is defined for $x\ne \frac{\pi }{2},\frac{3\pi }{2}$.

Now, $\tan x+\sec x=2\cos x$

$\begin{array}{rl}& \Rightarrow \frac{\sin x}{\cos x}+\frac{1}{\cos x}=2\cos x\\ & \Rightarrow (\sin x+1)=2{\cos }^{2}x\\ & \Rightarrow (\sin x+1)=2(1-{\sin }^{2}x)\\ & \Rightarrow (\sin x+1)=2(1-\sin x)(1+\sin x)\\ & \Rightarrow (1+\sin x)[2(1-\sin x)-1]=0\\ & \Rightarrow 2(1-\sin x)-1=0\\ & \dots \left[\begin{array}{r}\because \sin x\ne -1\text{ otherwise }\cos x=0\text{ and }\\ \tan x,\sec x\text{ will be undefined }\end{array}\right]\\ & \Rightarrow \sin x=\frac{1}{2}\\ & \Rightarrow x=\frac{\pi }{6},\frac{5\pi }{6}\text{ in }(0,2\pi )\end{array}$

$\therefore$ Number of solutions = 2

Correct answer option is (D)

$\begin{array}{rl}& {16}^{{\sin }^{2}x}+{16}^{{\cos }^{2}x}=10\\ & {16}^{{\sin }^{2}x}+{16}^{1-{\sin }^{2}x}=10\\ & {16}^{{\sin }^{2}x}+\frac{16}{{16}^{{\sin }^{2}x}}=10\end{array}$

Let ${16}^{{\sin }^{2}x}=t$

$\begin{array}{l}\therefore \mathrm{t}+\frac{16}{\mathrm{t}}=10\\ \therefore {\mathrm{t}}^2-10\mathrm{t}+16=0\\ \Rightarrow \mathrm{t}=2\text{ and }\mathrm{t}=8\end{array}$

Now, ${16}^{{\sin }^{2}x}=2$ and ${16}^{{\sin }^{2}x}=8$

$2^{4{\sin }^{2}x}=2^1\text{ and }{2}^{4{\sin }^{2}x}=2^3$

$\begin{array}{ll}\therefore & 4{\sin }^{2}x=1\text{ and }4{\sin }^{2}x=3\\ \therefore & {\sin }^{2}x=\frac{1}{4}\text{ and }{\sin }^{2}x=\frac{3}{4}\\ & \sin x=\pm \frac{1}{2}\text{ and }\sin x=\pm \frac{\sqrt{3}}{2}\\ \therefore & x=\frac{\pi }{6},\frac{5\pi }{6},\frac{7\pi }{6},\frac{11\pi }{6}\text{ and }x=\frac{\pi }{3},\frac{2\pi }{3},\frac{4\pi }{3},\frac{5\pi }{3}\end{array}$

$\therefore$ number of solutions $=8$.

Correct answer option is (A)

$\begin{array}{rl}& \cos 2\theta =\sin \theta \\ & \therefore 1-2{\sin }^2\theta =\sin \theta \Rightarrow 2{\sin }^2\theta +\sin \theta -1=0\\ & \therefore (2\sin \theta -1)(\sin \theta +1)=0\Rightarrow \sin \theta =\frac{1}{2},-1\end{array}$

When have $\theta \in (0,2\pi )$

$\therefore$ Possible values of $\theta$ are $\frac{\pi }{6},\frac{5\pi }{6},\frac{3\pi }{2}$

Correct answer option is (C)

We have $\tan \theta =\mathrm{k}\tan \varphi$ and $\theta +\varphi =\alpha$

$\therefore \frac{\tan \theta }{\tan \varphi }=\frac{\mathrm{k}}{1}$

By Componendo Dividendo, we get

$\begin{array}{rl}& \frac{\tan \theta +\tan \varphi }{\tan \theta -\tan \varphi }=\frac{\mathrm{k}+1}{\mathrm{k}-1}\\ & \therefore \frac{\frac{\sin \theta }{\cos \theta }+\frac{\sin \varphi }{\cos \varphi }}{\frac{\sin \theta }{\cos \theta }-\frac{\sin \varphi }{\cos \varphi }}=\frac{\mathrm{k}+1}{\mathrm{k}-1}\\ & \therefore \frac{\sin \cos \varphi +\cos \theta \sin \varphi }{\sin \theta \cos \varphi -\cos \theta \sin \varphi }=\frac{\mathrm{k}+1}{\mathrm{k}-1}\\ & \therefore \frac{\sin (\theta +\varphi )}{\sin (\theta -\varphi )}=\frac{\mathrm{k}+1}{\mathrm{k}-1}\Rightarrow \frac{\sin \alpha }{\sin (\theta -\varphi )}=\frac{\mathrm{k}+1}{\mathrm{k}-1}\\ & \therefore \sin (\theta -\varphi )=\frac{\mathrm{k}-1}{\mathrm{k}+1}(\sin \alpha )\end{array}$

Correct answer option is (B)

$\begin{array}{rl}& 2\sin (\theta +\frac{\pi }{3})=\cos (\theta -\frac{\pi }{6})\\ & 2[\sin \theta \cos \frac{\pi }{3}+\cos \theta \sin \frac{\pi }{3}]=(\cos \theta \cos \frac{\pi }{6}+\sin \theta \sin \frac{\pi }{6})\\ & \therefore 2(\frac{\sin \theta }{2}+\frac{\sqrt{3}\cos \theta }{2})=(\frac{\sqrt{3}\cos \theta }{2}+\frac{\sin \theta }{2})\\ & \therefore (\frac{\sin \theta }{2}+\frac{\sqrt{3}\cos \theta }{2})=0\Rightarrow \sin \theta +\sqrt{3}\cos \theta =0\Rightarrow \tan \theta =-\sqrt{3}\end{array}$